Let a ≠ b be two non-zero real numbers. Then the number of elements...

Given,

$X = \left\{z \in C: Re\left(a{z}^{2} + bz\right) = aandRe\left(b{z}^{2} + az\right) = b\right\}$

Now, let $z = x + iy$

So, $Re\left(a{z}^{2} + bz\right) = a$

$⇒ Re\left(a{\left(x + iy\right)}^{2} + b\left(x + iy\right)\right) = a$

$⇒ Re\left(a\left({x}^{2} - {y}^{2} + 2ixy\right) + b\left(x + iy\right)\right) = a$

$⇒ a\left({x}^{2} - {y}^{2}\right) + bx = a . . . \left(i\right)$

And, $Re\left(b{z}^{2} + az\right) = b$

$⇒ Re\left(b{\left(x + iy\right)}^{2} + a\left(x + iy\right)\right) = b$

$⇒ Re\left(b\left({x}^{2} - {y}^{2} + 2ixy\right) + a\left(x + iy\right)\right) = b$

$⇒ b\left({x}^{2} - {y}^{2}\right) + ax = b . . . \left(ii\right)$

From \left(i\right)$and \left(ii\right),$\left(i\right) - \left(ii\right)$we get,</p><p>\left({x}^{2} - {y}^{2}\right)$\left(a - b\right) - x\left(a - b\right) = a - b$</p><p>$⇒ {x}^{2} - {y}^{2} - x = 1 . . . \left(iii\right)$</p><p>From \left(i\right)$ and \left(ii\right), \left(i\right) + \left(ii\right)$</p><p>\left(\left({x}^{2} - {y}^{2}\right) + x - 1\right)\left(a + b\right) = 0$ (Note: here $a + b \ne 0$ is considered but it is not clear from the question)

$⇒ {x}^{2} - {y}^{2} + x - 1 = 0 . . . \left(iv\right)$

Now from \left(iii\right)$and \left(iv\right)$ we get,

$x = 0, {y}^{2} = - 1$ (No solution)